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Functional Analysis Solutions to Exercise Sheet 7

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Functional Analysis Solutions to Exercise Sheet 7 Institut für Analysis WS2019/20 Prof. Dr. Dorothee Frey 28.11.2019 M.Sc. Bas Nieraeth Functional Analysis Solutions to exercise sheet 7 Exercise 1: Strong convergence on compact sets. Let X and Y be Banach spaces. Suppose there is sequence of operators (Sn)n2N in L(Y ) such that Sn converges strongly to the identity operator on Y , i.e., for all y 2 Y we have lim kSny − yk = 0: n!1 (a) Show that for each compact set K ⊆ Y we have lim sup kSny − yk = 0: n!1 y2K (Hint: Find a suitable open cover of K. Also use the Banach-Steinhaus Theorem.) (b) Show that for each compact operator T 2 K(X; Y ) we have that SnT ! T as n ! 1 with respect to the operator norm. (c) Let p 2 [1; 1). Show that any compact operator T 2 K(X; `p) is the operator norm limit of a sequence of finite rank operators. Solution: (a) First note that since for each y 2 Y the sequence (Sny)n2N is convergent, this sequence is also bounded. Thus, by the Uniform Boundedness Principle we have supn2N kTnk < 1. " Let " > 0 and set "~ := sup kS k+2 . By our assumption on the Sn, for each y 2 Y we can n2N n find an Ny 2 N such that kSny − yk < "~ for n ≥ Ny: (1) Since K is compact, it is totally bounded. Thus, we can find a finite subset F ⊆ K such that (B(~y; "~))y~2F covers K. Now set N := maxy~2F Ny~. Let y 2 K and pick y~ 2 F such that y 2 B(~y; "~). Then if n ≥ N, we have n ≥ Ny~ so that by (1) we have kSny − yk ≤ kSny − Sny~k + kSny~ − y~k + ky~ − yk < kSnkky − y~k +" ~+" ~ < (sup kSnk + 2)~" = ": n2N Thus, whenever n ≥ N we have sup kSny − yk ≤ "; y2K proving the assertion. — Turn the page! — (b) As T is compact, the set T (BX ) is compact. Hence, by part (a), kSnT − T k = sup kSnT x − T xk = sup kSny − yk ≤ sup kSny − yk ! 0 kxk≤1 y2T (Bx) y2T (Bx) as n ! 1. p p (c) We define Sn : ` ! ` as the projection on the first n coordinates, i.e., Sny := (y1; : : : ; yn; 0; 0;:::). Then 1 0 1 1 p X p kSny − ykp = @ jyjj A ! 0 as n ! 1: j=n+1 p Now let T 2 K(X; ` ). Since SnT maps into the range of Sn, which is finite dimensional, SnT is a finite rank operator. Moreover, by part (b) we have limn!1 kSnT − T k = 0. We conclude that T is the operator norm limit of finite rank operators, as desired. p Exercise 2: Translations in L (R). (a) Let X be a Banach space and suppose we have a family of operators (T (t))t2R in L(X) satisfying the properties that T (s + t) = T (s)T (t) for all s; t 2 R and T (0) = Id. Show that the following are equivalent. (i) For each x 2 X the map R ! X, t 7! T (t)x is continuous. (ii) limt!0 T (t)x = x in X for all x 2 X; (iii) For every sequence (tn)n2N with tn ! 0 we have supn2N kT (tn)k < 1. Moreover, there is a dense subset D ⊆ X such that limt!0 T (t)x = x in X for all x 2 D. p p (b) Let p 2 [1; 1). For each t 2 R we define T (t): L (R) ! L (R) by (T (t)f)(x) = f(x + t). Show that the map p R ! L (R); t 7! T (t)f is continuous. You may use without proof that the space Cc(R) of continuous functions on R of compact support (i.e., the set of continuous functions f : R ! R such that p fx 2 R : f(x) 6= 0g is bounded) is dense in L (R). Solution: (a) We will prove (i),(ii) and (ii),(iii). “(i))(ii)” Since the map R ! X, t 7! T (t)x is, in particular, continuous at 0, we have, since T (0) = Id, that limt!0 T (t)x = T (0)x = x for all x 2 X. This proves (ii). “(ii))(i)” Fix t0 2 R. We will show that R ! X, t 7! T (t)x is continuous at t0. For x 2 X we have lim T (t)x = lim T (t − t0)T (t0)x = lim T (s)T (t0)x = T (t0)x t!t0 t!t0 s!0 by (ii). Thus, the map R ! X, t 7! T (t)x is continuous at every point, and hence, is continuous. “(ii))(iii)” Let (tn)n2N be a sequence converging to 0. Then by (ii) we have limn!1 T (tn)x = x for every x 2 X. In particular, the sequence (T (tn)x)n2N is bounded for each n 2 N. Hence, by the Uniform Boundedness Principle, we have supn2N kT (tn)k < 1. For the second assertion in (iii), we can set D = X. “(iii))(ii)” Let x 2 X. Note that the assertion limt!0 T (t)x = x is equivalent to the assertion that for every sequence (tn)n2N converging to 0, we have limn!1 T (tn)x = x. We prove this latter assertion. " Let (tn)n2N be a sequence with tn ! 0 and let " > 0. We set "~ = sup kT (t )k+2 and pick n2N n y 2 D such that kx − yk < "~ and N 2 N such that for all n ≥ N we have kT (tn)y − yk < "~. Then for all n ≥ N we have kT (tn)x − xk ≤ kT (tn)x − T (tn)yk + kT (tn)y − yk + ky − xk < kT (tn)kkx − yk +" ~+" ~ < (sup kT (tn)k + 2)~" = ": n2N Thus, limn!1 T (tn)x = x and the assertion follows. (b) First note that T (s + t) = T (s)T (t) for all s; t 2 R and T (0) = Id. Thus, to conclude p the result, it suffices to prove (iii) of part (a). First note that forall f 2 L (R) and t 2 R we have kT (t)fkp = kfkp by translation invariance of the Lebesgue measure. Hence, we have kT (t)k = 1 for all t 2 R and thus, in particular, for any sequence tn ! 0, supn2N kT (tn)k = 1. It remains to check the second condition of (iii). We set D := Cc(R). Let f 2 D and let r > 0 so that fx 2 R : f(x) 6= 0g ⊆ [−r; r]. We claim that f is uniformly continuous. For now, we assume the claim. Let " > 0. By uniform continuity of f we can pick δ~ > 0 such that if jtj < δ~, then p "p ~ jf(x + t) − f(x)j < 2r+2 for all x 2 R. Then set δ := minf1; δg. If jtj < δ, then jtj < δ~ and fx 2 R : f(x + t) 6= 0g ⊆ [−r − 1; r + 1]. Hence, Z Z r+1 p p p p " p kT (t)f − fkp = jf(x + t) − f(x)j dx = jf(x + t) − f(x)j dx ≤ (2r + 2) = " R −r−1 2r + 2 p so that kT (t)f − fkp ≤ ". Hence limt!0 T (t)f = f in L (R). The assertion follows. Finally, we prove the claim. Note that since the set [−r − 1; r + 1] is compact, the function f restricted to this set is uniformly continuous. Thus, for " > 0 we can pick δ~ > 0 such that whenever x; y 2 [−r − 1; r + 1] satisfy jx − yj < δ~, then jf(x) − f(y)j < ". Set δ := minf1; δg and let x; y 2 R such that jx − yj < δ. We consider two cases. In the first case, we assume that x; y 2 [−r − 1; r + 1]. Then, since jx − yj < δ ≤ δ~, we have jf(x) − f(y)j < ", as desired. In the other case we have x2 = [−r − 1; r + 1] or — Turn the page! — y2 = [−r − 1; r + 1]. We assume without loss of generality that x2 = [−r − 1; r + 1]. But then, since jx − yj < δ ≤ 1, we have y2 = [−r; r]. This means that f(x) = f(y) = 0 so that jf(x) − f(y)j = 0 < ", as desired. We conclude that f is indeed uniformly continuous, proving the claim. Exercise 3: Duality in sequence spaces. Let y = (yj)j2N be a sequence in K. (a) Suppose that 1 X xjyj j=1 1 converges in K for all (xj)j2N 2 c0. Show that y 2 ` . (Hint: use the Banach-Steinhaus Theorem.) (b) Let p 2 [1; 1]. Suppose that 1 X xjyj j=1 p p0 1 1 converges in K for all (xj)j2N 2 ` . Show that y 2 ` , where p + p0 = 1. Solution: Pn (a) For x 2 c0 we define Tnx := j=1 xjyj. Then Tn is linear and we have n n X X jTnxj ≤ jxjyjj ≤ kxk1 jyjj; j=1 j=1 showing that Tn 2 L(c0; K). Moreover, we assumed that (Tnx)n2N is a convergent sequence in K for all x 2 c0. Thus, by the Banach-Steinhaus Theorem we have supn2N kTnk < 1. −1 Fix J 2 N. we define a sequence x 2 c0 by xj = yjjyjj (with the understanding that xj = 0 if yj = 0) for j 2 f1;:::;Jg and xj = 0 for j > J. Then, kxk1 ≤ 1 and J X jyjj = jTJ xj ≤ kTJ k ≤ sup kTnk j=1 n2N 1 Taking a supremum over J 2 N proves that y 2 ` , as desired.
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